# Question #fbf2a

Jun 28, 2017

This question may be missing some important info to proceed, so I made it up along the way. For my made-up volume of gastric acid you need about 4.5 tablets. The dosage would be 5, probably.

#### Explanation:

Let's first obtain the HCl molarity in the gastric "juice" solution.

$2.19 g \left(\frac{H C l}{36.463 g}\right) = 0.0601 m o l = \frac{0.0601 m o l}{L} = 0.0601 M$

This is the balanced neutralization reaction.

$3 H C l \left(a q\right) + A l {\left(O H\right)}_{3} \left(s\right) \to A l C {l}_{3} \left(a q\right) + 3 {H}_{2} O \left(l\right)$

Normally, there is some amount of solution to neutralize, here evidently none is given. I think this is almost needed in order to proceed (please correct me in the comments if not!), so I will make one up for example's sake. Let's say we have 1.50L of gastric "juice", a reasonable estimate.

$1.50 L \left(\frac{0.0601 m o l}{L}\right) \left(\frac{A l {\left(O H\right)}_{3}}{3 H C l}\right) \left(\frac{78.01 g}{A l {\left(O H\right)}_{3}}\right) \left(\frac{{10}^{3} m g}{g}\right) = 2.34 \cdot {10}^{3} m g$

Lastly, we'd then divide this mass of $A l {\left(O H\right)}_{3}$ by the mass of one antacid tablet to arrive at our answer.

$\frac{2.34 \cdot {10}^{3} m g}{520 m g} = 4.5$ tablets